用单片机配置FPGA—PLD设计技巧 Configuration/Program Method for Altera Device Configure the FLEX Device You can use any Micro-Controller to configure the FLEX device–the main idea is clocking in ONE BITof configuration data per CLOCK–start from the BIT 0􀂄The total Configuration time–e.g. 10K10 need 15K byte configuration file•calculation equation–10K10* 1.5= 15Kbyte–configuration time for the file itself•15*1024*8*clock = 122,880Clock•assume the CLOCK is 4MHz•122,880*1/4Mhz=30.72msec
上传时间： 2013-10-09
上传用户：a67818601
|Introduction􀂄􀂄 Basic Concept􀂄􀂄 Tips to layout Power circuit􀂄􀂄 Type of Power circuit Basic Concept􀂄􀂄 Maximum Current calculation􀂄􀂄 Resistance of Copper􀂄􀂄 ideal power supply & noise􀂄􀂄 Capacitor & Inductor􀂄􀂄 Power consumption􀂄􀂄 Function of power circuit
上传时间： 2013-12-09
上传用户：JIEWENYU
图像识别技术目前在人们的生活中，应用越来越普遍，如我们较熟悉的数码相机、摄像头、具有摄像功能的手机、以及我们看见的一些智能玩具上都具有图像识别的功能。在此为大家介绍的此款图像识别模组主要是应用于交互式智能玩具、图像处理产品、教学中。此模组接口简单，应用方便，容易与MCU结合使用，若与凌阳公司的SPCE061A结合，使其声貌并具，更能体现产品的特效功能。 2 功能简介 该模组可以实现如下功能: 􀂄 识别颜色、形状 􀂄 识别位置 3 模组结构 光学镜头CMOS传感器SPCA561A图象处理芯片SPCA563A微控制器SPCE061AEagle模组 图3.1 图像识别模组结构框图 图像识别模组由光学镜头、CMOS传感器（SPCA561A）、图像处理芯片（SPCA563A）组成。 SPCA563A分析和处理SPCA561A传过来的图像信号，并得出相应的信息（颜色、形状等等）。此模组接口简单，应用方便，容易与MCU结合使用，我们下面将要介绍的Demo就是其与SPCE061A结合的例子。 图像识别模组和外界的接口共有6针，依次接在SPCA563A的：VCC、SCK、SD、RDY、图像识别3_RESET和GND。
标签： 图像识别
上传时间： 2013-12-12
上传用户：talenthn
The basic principle using the branchand- bound strategy to solve the traveling salesperson optimization problem (TSP) consists of two parts. 􀂄 There is a way to split the solution space. 􀂄 There is a way to predict a lower bound for a class of solutions. 􀂄 There is also a way to find an upper bound of an optimal solution. 􀂄 If the lower bound of a solution exceeds this upper bound, this solution cannot be optimal. Thus, we should terminate the branching associated with this solution.
标签： salesperson principle branchand the
上传时间： 2017-02-18
上传用户：comua
Instead of finding the longest common subsequence, let us try to determine the length of the LCS. 􀂄 Then tracking back to find the LCS. 􀂄 Consider a1a2…am and b1b2…bn. 􀂄 Case 1: am=bn. The LCS must contain am, we have to find the LCS of a1a2…am-1 and b1b2…bn-1. 􀂄 Case 2: am≠bn. Wehave to find the LCS of a1a2…am-1 and b1b2…bn, and a1a2…am and b b b b1b2…bn-1 Let A = a1 a2 … am and B = b1 b2 … bn 􀂄 Let Li j denote the length of the longest i,g g common subsequence of a1 a2 … ai and b1 b2 … bj. 􀂄 Li,j = Li-1,j-1 + 1 if ai=bj max{ L L } a≠b i-1,j, i,j-1 if ai≠j L0,0 = L0,j = Li,0 = 0 for 1≤i≤m, 1≤j≤n.
标签： the subsequence determine Instead
上传时间： 2013-12-17
上传用户：evil
